3. Решите уравнение: a) $$3x^2 = 2x + 4$$; б) $$(x - 1)(2x + 3) = -2$$; в) $$\frac{x^2 + 7}{2} = 4x$$.

a) $$3x^2 = 2x + 4 Rightarrow 3x^2 - 2x - 4 = 0$$ $$D = (-2)^2 - 4(3)(-4) = 4 + 48 = 52$$ $$x_1 = \frac{2 + \sqrt{52}}{6} = \frac{2 + 2\sqrt{13}}{6} = \frac{1 + \sqrt{13}}{3}$$ $$x_2 = \frac{2 - \sqrt{52}}{6} = \frac{2 - 2\sqrt{13}}{6} = \frac{1 - \sqrt{13}}{3}$$ б) $$(x - 1)(2x + 3) = -2 Rightarrow 2x^2 + 3x - 2x - 3 = -2 Rightarrow 2x^2 + x - 1 = 0$$ $$D = 1^2 - 4(2)(-1) = 1 + 8 = 9$$ $$x_1 = \frac{-1 + \sqrt{9}}{4} = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ $$x_2 = \frac{-1 - \sqrt{9}}{4} = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ в) $$\frac{x^2 + 7}{2} = 4x Rightarrow x^2 + 7 = 8x Rightarrow x^2 - 8x + 7 = 0$$ $$D = (-8)^2 - 4(1)(7) = 64 - 28 = 36$$ $$x_1 = \frac{8 + \sqrt{36}}{2} = \frac{8 + 6}{2} = \frac{14}{2} = 7$$ $$x_2 = \frac{8 - \sqrt{36}}{2} = \frac{8 - 6}{2} = \frac{2}{2} = 1$$